how to defer executing argument in defer function

Question

defer in go schedules a function call to be run after the function completes. So 1st print function fmt.Println("a ", Even()) is defered. But as argument is executed when defer is called, Even() function is called immediately.

package main

import "fmt"

func MakeEvenNumber() func() int {
    num :=0
    return func() (ret int) {
        ret = num
        num = num + 2
        return ret
    }
}

func main() {

    Even := MakeEvenNumber()

    defer fmt.Println("a ", Even())

    fmt.Println("b ", Even())
    fmt.Println("c ", Even())
    fmt.Println("d ", Even())
}

Output:

b  2
c  4
d  6
a  0

This is happened in defer. I just want to know any way to stop executing argument in defer function?

like in this case, output will be:

b 0
c 2
d 4
a 6

defer fmt.Println("a ", **Even()**) this Even() function will be executed along with Println.

Any way to do this?

score 3 · Accepted Answer · answered Aug 23 '15 at 14:58

If you wrap the fmt.Println call in another function (an anonymous function, in this case), the evaluation of Even() will happen after the other calls to Even have completed:

func main() {
    Even := MakeEvenNumber()

    defer func() {
      fmt.Println("a ", Even())
    }()

    fmt.Println("b ", Even())
    fmt.Println("c ", Even())
    fmt.Println("d ", Even())
}

how to defer executing argument in defer function

1 Answers1